Do I have a problem? Basic proportion problem.

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Homework Help Overview

The discussion revolves around a basic proportion problem related to a math homework assignment, specifically focusing on the interpretation of the problem and the methods used to solve it.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to compare their method of solving a proportion problem with the solution provided in a textbook, expressing concern about the complexity of their approach. Some participants question the clarity of the problem statement and its interpretation, particularly regarding the number of questions completed versus remaining.

Discussion Status

The discussion includes various interpretations of the problem setup, with participants offering different approaches to arrive at the same answer. There is a recognition that both methods yield the correct result, and some guidance is provided regarding the validity of using proportions in linear problems.

Contextual Notes

Participants note the potential for misunderstanding in the problem statement, particularly regarding the phrasing of how many questions have been completed versus how many are left to complete. The original poster expresses concern about the implications of their approach for future studies in engineering.

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Hello, this is an odd question but when I compared how I solved it compared to how the book solved it I kinda got worried. It was a very basic problem I been going back and touching up on my basics before I start pre-cal and trig in college.

The problem was: After completing 7/10 of his math homework assignment, Josh has 15 more questions complete. What is the total number of questions on his assignment? The answer was 50.

I got that by

[itex]\frac{7}{10}[/itex]=[itex]\frac{(x-15)}{x}[/itex] Followed by cross multiplying and dividing.

However the book does it simply by doing the subtraction 10-7 = 3 then setting it up as

[itex]\frac{3}{10}[/itex]=[itex]\frac{15}{x}[/itex] Then cross multiply and divide.

I know someone will say "It works so you can do what you are doing". But is there a possibility this kind of over complication( haha not hardly complicated!) Can hurt me later on? The way I performed it seemed much more intuitive... I have a high school record of up to ap calc and ap chem and I'm worried if I pursue engineering in college I'll have a hard time solving simple things because of my use to complexity?
 
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(3/10)x=15
x=50
 
Life-Like said:
Hello, this is an odd question but when I compared how I solved it compared to how the book solved it I kinda got worried. It was a very basic problem I been going back and touching up on my basics before I start pre-cal and trig in college.

The problem was: After completing 7/10 of his math homework assignment, Josh has 15 more questions complete.
Do you meant "15 more questions to complete"? What you wrote could be interpreted as meaning has completed 15 questions. If he has already completed 7/10 of his assignment, he still has 3/10 left. If he has 15 questions more to complete, letting "x" be the total number of problems, (3/10)x= 15 so x= (10/3)(15)= 150/3= 50.

What is the total number of questions on his assignment? The answer was 50.

I got that by

[itex]\frac{7}{10}[/itex]=[itex]\frac{(x-15)}{x}[/itex] Followed by cross multiplying and dividing.

However the book does it simply by doing the subtraction 10-7 = 3
Okay, that is, they are calculating 1- 7/10= 10/10- 7/10= (10- 7)/10= 3/10 as I did.

then setting it up as

[itex]\frac{3}{10}[/itex]=[itex]\frac{15}{x}[/itex] Then cross multiply and divide.

I know someone will say "It works so you can do what you are doing". But is there a possibility this kind of over complication( haha not hardly complicated!) Can hurt me later on? The way I performed it seemed much more intuitive... I have a high school record of up to ap calc and ap chem and I'm worried if I pursue engineering in college I'll have a hard time solving simple things because of my use to complexity?
As long as you have linear problems, you can set them up as "proportions" and get the correct answer.

 
Thanks guys
 
Last edited:

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